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User blog:Alemagno12/yet another BMS analysis
Since BMS is becoming a popular topic now, why not have another try at an analysis? (This'll probably be split into parts if it gets long enough) Up to ε0 The blank sequence corresponds to 0. Then, adding (0)s increases the ordinal by 1. *(0) = 1 *(0)(0) = 2 *(0)(0)(0) = 3 (1) takes the limit of duplicating the (0) to its left. *(0)(1) = (0)(0)(0)... = ω *(0)(1)(0) = ω+1 *(0)(1)(0)(0) = ω+2 *(0)(1)(0)(1) = (0)(1)(0)(0)(0)... = ω2 *(0)(1)(0)(1)(0) = ω2+1 *(0)(1)(0)(1)(0)(1) = ω3 *(0)(1)(0)(1)(0)(1)(0)(1) = ω4 More specifically, (1) takes the limit of duplicating the (0)-type separator group to its left, which is just the rightmost (0) to its left and all elements to the right of that, excluding, of course, the (1). *(0)(1)(1) = (0)(1)(0)(1)... = ω2 *(0)(1)(1)(0) = ω2+1 *(0)(1)(1)(0)(1) = ω2+ω *(0)(1)(1)(0)(1)(0)(1) = ω2+ω2 *(0)(1)(1)(0)(1)(1) = (0)(1)(1)(0)(1)(0)(1)... = ω22 *(0)(1)(1)(0)(1)(1)(0)(1) = ω22+ω *(0)(1)(1)(0)(1)(1)(0)(1)(1) = ω23 *(0)(1)(1)(1) = (0)(1)(1)(0)(1)(1)... = ω3 *(0)(1)(1)(1)(0)(1)(1) = ω3+ω2 *(0)(1)(1)(1)(0)(1)(1)(1) = ω32 *(0)(1)(1)(1)(1) = (0)(1)(1)(1)(0)(1)(1)(1)... = ω4 *(0)(1)(1)(1)(1)(1) = ω5 *(0)(1)(1)(1)(1)(1)(1) = ω6 In general, (a) takes the limit of duplicating the (a-1)-type separator group to its left, which is just the rightmost (a-1) to its left and all elements to the right of that, excluding, of course, the (a). *(0)(1)(2) = (0)(1)(1)(1)... = ωω *(0)(1)(2)(0) = ωω+1 *(0)(1)(2)(0)(1) = ωω+ω *(0)(1)(2)(0)(1)(2) = ωω2 *(0)(1)(2)(1) = ωω+1 *(0)(1)(2)(1)(2) = (0)(1)(2)(1)(1)... = ωω2 *(0)(1)(2)(1)(2)(1)(2) = ωω3 *(0)(1)(2)(2) = (0)(1)(2)(1)(2)... = ωω2 *(0)(1)(2)(2)(1)(2)(2) = ωω22 *(0)(1)(2)(2)(2) = (0)(1)(2)(2)(1)(2)(2)... = ωω3 *(0)(1)(2)(2)(2)(2) = ωω4 *(0)(1)(2)(3) = (0)(1)(2)(2)(2)... = ωωω *(0)(1)(2)(3)(2) = ωωω+1 *(0)(1)(2)(3)(2)(3) = ωωω2 *(0)(1)(2)(3)(3) = (0)(1)(2)(3)(2)(3)... = ωωω2 *(0)(1)(2)(3)(3)(3) = ωωω3 *(0)(1)(2)(3)(4) = (0)(1)(2)(3)(3)(3)... = ωωωω *(0)(1)(2)(3)(4)(4) = ωωωω2 *(0)(1)(2)(3)(4)(5) = ωωωωω *(0)(1)(2)(3)(4)(5)(6) = ω↑↑6 *(0)(1)(2)(3)(4)(5)(6)(7) = ω↑↑7 So the limit of Primitive Sequence System is ε0. You might've noticed that it works as an encoding for Cantor's normal form: concatenations of(0)-type separator groups encode sums of ordinals, concatenations of (1)-type separator groups encode sums of ordinals in the exponent of an ω, concatenations of (2)-type separator groups encode sums of ordinals in the exponent of an ω in the exponent of another ω, and so on Up to εω All elements we have seen so far, which are of the form (a), are zeroth-level elements, and so they can be expressed as (a,0). We introduce the first first-level element, which is (1,1). It works like (1,0), but at every duplication, the first entry is increased by one, until we reach the limit of the duplications. *(0,0)(1,1) = (0,0)(1,0)(2,0)... = ε0 *(0,0)(1,1)(0,0) = ε0+1 *(0,0)(1,1)(0,0)(1,0) = ε0+ω *(0,0)(1,1)(0,0)(1,0)(2,0) = ε0+ωω *(0,0)(1,1)(0,0)(1,1) = ε02 *(0,0)(1,1)(0,0)(1,1)(0,0)(1,1) = ε03 *(0,0)(1,1)(1,0) = ωε0+1 *(0,0)(1,1)(1,0)(0,0)(1,1)(1,0) = ωε0+12 *(0,0)(1,1)(1,0)(1,0) = ωε0+2 *(0,0)(1,1)(1,0)(2,0) = ωε0+ω *(0,0)(1,1)(1,0)(2,0)(3,0) = ωε0+ωω *(0,0)(1,1)(1,0)(2,0)(3,0)(4,0) = ωε0+ωωω And, in general, (a,1) works like (a,0), but at every duplication, the first entry is increased by one, until we reach the limit of the duplications. *(0,0)(1,1)(1,0)(2,1) = (0,0)(1,1)(1,0)(2,0)(3,0)... = ωε02 *(0,0)(1,1)(1,0)(2,1)(1,0) = ωε02+1 *(0,0)(1,1)(1,0)(2,1)(1,0)(2,1) = ωε03 *(0,0)(1,1)(1,0)(2,1)(2,0) = ωωε0+1 *(0,0)(1,1)(1,0)(2,1)(2,0)(2,0) = ωωε0+2 *(0,0)(1,1)(1,0)(2,1)(2,0)(3,0) = ωωε0+ω *(0,0)(1,1)(1,0)(2,1)(2,0)(3,0)(4,0) = ωωε0+ωω *(0,0)(1,1)(1,0)(2,1)(2,0)(3,1) = (0,0)(1,1)(1,0)(2,1)(2,0)(3,0)... = ωωε02 *(0,0)(1,1)(1,0)(2,1)(2,0)(3,1)(3,0) = ωωωε0+1 *(0,0)(1,1)(1,0)(2,1)(2,0)(3,1)(3,0)(4,0) = ωωωε0+ω *(0,0)(1,1)(1,0)(2,1)(2,0)(3,1)(3,0)(4,1) = (0,0)(1,1)(1,0)(2,1)(2,0)(3,1)(3,0)(4,0)... = ωωωε02 *(0,0)(1,1)(1,0)(2,1)(2,0)(3,1)(3,0)(4,1)(5,0)(6,1) = ωωωωε02 *(0,0)(1,1)(1,0)(2,1)(2,0)(3,1)(3,0)(4,1)(5,0)(6,1)(6,0)(7,1) = ωωωωωε02 Following the Cantor normal form encoding, adding a (1,1) is the same as getting to the next fixed point of ω exponentiation, i.e the next epsilon number. *(0,0)(1,1)(1,1) = (0,0)(1,1)(1,0)(2,1)(2,0)(3,1)... = ε1 *(0,0)(1,1)(1,1)(0,0)(1,1)(1,1) = ε12 *(0,0)(1,1)(1,1)(1,0) = ωε1+1 *(0,0)(1,1)(1,1)(1,0)(2,1) = ωε1+ε0 *(0,0)(1,1)(1,1)(1,0)(2,1)(2,1) = (0,0)(1,1)(1,1)(1,0)(2,1)(2,0)(3,1)... = ωε12 *(0,0)(1,1)(1,1)(1,0)(2,1)(2,1)(2,0) = ωωε1+1 *(0,0)(1,1)(1,1)(1,0)(2,1)(2,1)(2,0)(3,1)(3,1) = (0,0)(1,1)(1,1)(1,0)(2,1)(2,1)(3,0)(4,1)(4,0)(3,1)... ωωε12 *(0,0)(1,1)(1,1)(1,0)(2,1)(2,1)(2,0)(3,1)(3,1)(3,0)(4,1)(4,1) = ωωωε12 *(0,0)(1,1)(1,1)(1,1) = (0,0)(1,1)(1,1)(1,0)(2,1)(2,1)... = ε2 *(0,0)(1,1)(1,1)(1,1)(1,0) = ωε2+1 *(0,0)(1,1)(1,1)(1,1)(1,0)(2,1)(2,1)(2,1) = (0,0)(1,1)(1,1)(1,1)(1,0)(2,1)(2,1)(2,0)(3,1)(3,1)... = ωε22 *(0,0)(1,1)(1,1)(1,1)(1,0)(2,1)(2,1)(2,1)(2,0)(3,1)(3,1)(3,1) = ωωε22 *(0,0)(1,1)(1,1)(1,1)(1,1) = (0,0)(1,1)(1,1)(1,1)(1,0)(2,1)(2,1)(2,1)... = ε3 *(0,0)(1,1)(1,1)(1,1)(1,1)(1,0)(2,1)(2,1)(2,1)(2,1) = ωε32 *(0,0)(1,1)(1,1)(1,1)(1,1)(1,1) = (0,0)(1,1)(1,1)(1,1)(1,1)(1,0)(2,1)(2,1)(2,1)(2,1)... = ε4 *(0,0)(1,1)(1,1)(1,1)(1,1)(1,1)(1,1) = ε5 *(0,0)(1,1)(1,1)(1,1)(1,1)(1,1)(1,1)(1,1) = ε6 Up to φ(ω,0)? (x,0)s can not only diagonalize over duplications of (x-1,0)-type separator groups, but also of (x-1,a)-type separator groups for any a. *(0,0)(1,1)(2,0) = (0,0)(1,1)(1,1)(1,1)... = εω *(0,0)(1,1)(2,0)(0,0) = εω+1 *(0,0)(1,1)(2,0)(0,0)(1,1)(2,0) = εω2 *(0,0)(1,1)(2,0)(1,0) = ωεω+1 *(0,0)(1,1)(2,0)(1,0)(2,1)(3,0) = (0,0)(1,1)(2,0)(1,0)(2,1)(2,1)(2,1)... = ωεω2 *(0,0)(1,1)(2,0)(1,0)(2,1)(3,0)(2,0)(3,1)(4,0) = ωωεω2 *(0,0)(1,1)(2,0)(1,1) = εω+1 *(0,0)(1,1)(2,0)(1,1)(1,0)(2,1)(3,0)(2,1) = ωεω+12 *(0,0)(1,1)(2,0)(1,1)(1,1) = εω+2 *(0,0)(1,1)(2,0)(1,1)(2,0) = (0,0)(1,1)(2,0)(1,1)(1,1)(1,1)... εω2 *(0,0)(1,1)(2,0)(1,1)(2,0)(1,1)(2,0) = εω3 *(0,0)(1,1)(2,0)(2,0) = (0,0)(1,1)(2,0)(1,1)(2,0)(1,1)(2,0)... = εω2 *(0,0)(1,1)(2,0)(2,0)(1,1)(2,0)(2,0) = εω22 *(0,0)(1,1)(2,0)(2,0)(2,0) = (0,0)(1,1)(2,0)(2,0)(1,1)(2,0)(2,0)... = εω3 *(0,0)(1,1)(2,0)(3,0) = εωω *(0,0)(1,1)(2,0)(3,0)(4,0) = εωωω *(0,0)(1,1)(2,0)(3,1) = εε0 *(0,0)(1,1)(2,0)(3,1)(3,1) = εε1 *(0,0)(1,1)(2,0)(3,1)(4,0) = (0,0)(1,1)(2,0)(3,1)(3,1)(3,1)... = εεω *(0,0)(1,1)(2,0)(3,1)(4,0)(5,1) = εεε0 *(0,0)(1,1)(2,0)(3,1)(4,0)(5,1)(6,0) = εεεω *(0,0)(1,1)(2,0)(3,1)(4,0)(5,1)(6,0)(7,1) = εεεε0 *(0,0)(1,1)(2,0)(3,1)(4,0)(5,1)(6,0)(7,1)(8,0)(8,1) = εεεεε0 WIP Category:Blog posts